no tagsLecture 007: Applications of Gauss's Law and ConductorsSteveSekula, 8 September 2010 (created 2 September 2010)GoalsLearn to work problems involving Gauss's LawWorking with Gauss's LawAs in all problems in physics, all the work is in setting up the question. Ifyou can narrow down on what's being asked, and how to translate that intothe language of mathematics (sprinkling in some physics thinking for goodmeasure), then you will be able to setup and solve hard problems. I'll worktwo problems today to illustrate this.Applying Gauss's Law: A Uniformly Charged SphereImagine that you construct a spherical structure of charge, where thecharge is uniformly distributed throughout the volume, whose radius is .Can you find the electric field at all points?To attack this problem, we can first consider the geometry. We have somecharge spread uniformly throughout this sphere. That means we have aconstant volume charge density, . Independent of the sub-volume, , ofthe sphere we consider, will be the same. One can alreadysuspect that this will come in handy for rewriting charge in termed ofsomething we can actually integrate.There are really two "regions of interest" in this problem: points inside thesphere, and points outside the sphere. The distribution has sphericalsymmetry, and so we should be able to "easily" use Gauss's Law to solve forthe field.General Physics - E&M (PHY 1308) LectureNotesGeneral Physics - E&M (PHY 1308) LectureNotesR Q Ú dV Ú q=dV = dGeneral Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...1 of 6 09/08/2010 10:15 AMGauss's Law works best when you have symmetry of some kind: spherical,planar, linear.Let's begin by writing down Gauss's Law:Let's then consider two Gaussian Surfaces: a spherical surface inside thecharged sphere, with its own radius ( ), and a spherical surface thatencloses the entire charge sphere and with . Just given the symmetryof this sphere, you can already see that whatever the electric field, itsvectors will point radially outward from the center of the charge-sphere.We need to attack that flux integral on the left-hand side. We can use whatwe have just observed about the charge-sphere to start simplifying things:First, given the spherical symmetry of the problem, we already knowthat is parallel to the normal pointing out from our Gaussian surfacepiece, . Thus in the dot product, since. So the flux integral just boils down to .With our Gaussian surface centered on the charge sphere, it's the samedistance from all points along the same radius on the enclosed charge -therefore whatever the field, its value is constant across the Gaussiansurface and thus can come out of the integral, .The integral is going to simply yield the total surface area of asphere of radius , the radius of our Gaussian surface. Thus.And we've evaluated the surface integral.Now we need to evaluate the right-hand side of Gauss's Law: . Weneed to determine the amount of charge enclosed by our Gaussian surface.It will matter whether or . For , the answer is straight-forward: , the whole charge of the sphere. What about for ?È A =Ï =ZsurfaceE~Á d~= qenclosed0r r < Rr > RE ~dA ~E A dA dA ~Á d~= E cos Ò = Ecos Ò = 1dA RsurfaceEÈ A = ERsurfacedA Rsurfacedr È A (4Ùr ) = E = E2q =Ï enclosed0r < R r > R r > Rq enclosed= Q r < RGeneral Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...2 of 6 09/08/2010 10:15 AMTo tackle this, we need to take advantage of the constant volumedensity of charge; that is, is the same regardless of the volume insidethe sphere we choose to evaluate. Let's consider two different ways ofobtaining . We could consider the whole volume of the charge-sphere.In that case,We could also consider the Gaussian sphere for which . In thatcase, [rho(r) = rho = q_{enclosed}/((4/3) pi r^3).] These two s areequivalent, so setting them equal we find thatLet's put all of these pieces together - the right-hand side of Gauss's Lawand the left-hand side of Gauss's Law:For :This reduces toE(r )For :which reduces toE(r ) :Draw the function on the board, with the linear piece for and theinverse-square law piece for .Does this make sense? Inside the sphere, as we grow the size of ourGaussian surface, we include more and more charge and we expect thefield strength to grow. Outside the sphere, we fully enclose all the chargeand just get further from the charge as gets bigger, so we expect this toÚ Ú Ú(R) =((4=3)ÙR ): = Ú = Q3r < RÚ Q=((4=3)ÙR ) =(4=3)Ùr ) 3= qenclosed3! qenclosed= Q°rRÑ3:r < RA =Ï (4Ùr ) (r=R) (1=Ï ): ZsurfaceE~Á d~= qenclosed0! E2= Q30< R =Qr4ÙÏ R03r > RE(4Ùr ) =Ï 2= Q0> R =Q4ÙÏ r02r < Rr > Rr General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...3 of 6 09/08/2010 10:15 AMconverge at large distances to the electric field of a point charge, , whichit does.Imagine how hard it would be to calculate this using just the superpositionprinciple! It's a very beautiful and useful result. Incidentally, this law andthis solution holds for gravity as well. It's why we can treat planets as"point masses" when were are outside of the planet's surface (as if all themass were concentrated at the center of the planet in a point).But what if our charge is NOT uniformly distributed in the interior?Applying Gauss's Law: A Thin, Hollow Spherical ShellImagine a thin spherical shell of charge, with an outer radius . It's thinenough that we can neglect its thickness for this problem. What is theelectric field inside and outside the shell?Again, this is a spherically symmetric distribution. Outside the shell, all ofthe charge is enclosed again an the field outside looks just like it did forthe solid sphere of charge. We just have to concern ourselves with the fieldINSIDE.Construct a spherical Gaussian surface with radius centered onthe shell's center. Again, for all the same reasons as in the chargedsphere the flux is just .We're left to just evaluate the right-hand side of Gauss's Law. Thegaussian surface encloses no charge, so .Therefore, everywhere inside the shell!Due to the inverse square law, at any point inside the shell the fields fromcharges on the side of the shell closer to the point are exactly cancelled byall of the fields from charges on the side further away. This is a remarkablefact, one which can be tested through experiment (we'll discuss this
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